22.314/1.56/2.084/13.14
Fall 2006
Problem Set II Solution
1. Stress intensity = max{|σr − σθ |, |σθ − σz |, σz − σr |}
For thin wall approximation:
Pi + Po
2
Pi R2 − Po (R + t)2
σz = −
(R + t)2 − R2
Pi − P o
t
σθ =
(R + )
t
2
σr = −
(1)
(2)
(3)
Therefore:
Sthin = σθ − σr =
Pi − P o
t
Pi + Po
(R + ) +
t
2
2
(4)
Thick wall solution:
Equilibrium in radial direction gives:
dσr σr − σθ
+
=0
dr
r
(5)
Hook’s law:
1
(σr − νσθ − νσz )
E
1
�θ = (σθ − νσr − νσz )
E
1
�z = (σz − νσr − νσθ )
E
�r =
Since �θ = u/r, �r =
du
,
dr
(6)
(7)
(8)
we get:
d�θ
1
= (�r − �θ )
dr
r
(9)
For this close end cylinder far from the end, plane stress condition is assumed, i.e., σz is
const.
Plug Eq 6 and Eq 5into Eq 9, we get
d
(σθ + σr ) = 0
dr
(10)
Plug Eq 10 into Eq 5, we get
d 1 d 2
(r σr ) = 0
dr r dr
(11)
1
With B.C. σr (r = R) = −Pi and σr (r = R + t) = −Po , we get:
R
R −Po (R + t)2 + Pi R2
σr = −Pi ( )2 + (1 − ( )2 )
(R + t)2 − R2
r
r
R
R −Po (R + t)2 + Pi R2
σθ = Pi ( )2 + (1 + ( )2 )
(R + t)2 − R2
r
r
R (Pi − Po )(R + t)2
σθ − σr = 2( )2
(R + t)2 − R2
r
(12)
(13)
(14)
Maximum stress intensity is at the location of inner radius:
Sthick = 2
(Pi − Po )(R + t)2
(R + t)2 − R2
The error in thin wall approximation is:
|1 −
Sthin
|
Sthick
Results are tabulated below:
t/R
Pi = 2Po
Pi = 20Po
0.03
1.41%
1.31%
0.10
0.15
4.13% 5.67%
4.09% 5.88%
0.30
8.88%
10.46%
2. We use thin shell approximation to solve this problem. For a region of cylinder far from a
junction, stresses are:
PR
t
σr = −P/2
PR
σx =
2t
σθ =
(15)
(16)
(17)
Radial displacement:
uc =
P R2
νt
(2 − ν + )
R
2Et
(18)
For the sphere, stresses are:
PR
2t
σr = −P/2
σθ = σφ =
(19)
(20)
Radial displacement:
us =
P R2
νt
(1 − ν + )
2Et
R
(21)
2
At the junction, Note C Eq 30–33 give that at the edge of cylinder:
u0 = u c +
φ0 = −
V0
M0
+
2β 3 D 2β 2 D
(22)
V0
M0
−
2β 2 D βD
(23)
at the edge of hemisphere:
u0 = u s −
φ0 = −
2Rλ
2λ2
V0 +
M0
Et
Et
(24)
2λ2
4λ3
V0 +
M0
Et
REt
(25)
2
where, λ = βs R, βs = ( 3(1−ν
)1/4 , u0 is the radial displacement at the junction, φ0 is the
R2 t2
1
slope at the junction.
Due to the continuity of the displacement and slope, the four unknowns u0 , φ0 , M0 , and V0
1
Note that the direction of the shear force Q in Note L4 is different from that in Note C. If you assume the
direction of Q0S is same as QOC , the continuity of shear force should give that: Q0S = −Q0C .
Vo
Mo
Mo
z
Vo
x
Figure 1: Junction of hemisphere and cylinder
3
can be solved by above equations. It can be found that
M0 = 0
V0 = −P R2 /(4Rλ + Et/β 3 D)
V0
u = uc +
2Dβ 3
P R
σx =
2t
PR
EV0
σθ =
+
t
2DRβ 3
σr = −P/2
(26)
(27)
(28)
(29)
(30)
(31)
With mean radius R = inner radius + t/2 = 1.155 m, t = 0.11 m, E = 200 GPa, ν = 0.3, we
get
(a) At the junction, from Eq 29–Eq 31:
σx = 81.38 MPa
σθ = 122.06 MPa
σr = -7.75 MPa
The maximum stress is the hoop stress: 122.06 MPa.
(b) Radial displacement as a function of radial postion z is: (R + z)�θ . Thus, from Eq 18,
the radial displacement of cylinder is:
PR
νt
(2 − ν + )(R + z)
2Et
R
From Eq 21, the radial displacement of hemishpere is:
PR
νt
(1 − ν + )(R + z)
2Et
R
From Eq 28, the radial displacement of junction is:
(
PR
νt
V0
(2 − ν + ) +
)(R + z)
2Et
R
2DRβ 3
Therefore:
Max. displacement (m)
Cylinder Sphere
0.00089 0.00037
4
Junction
0.00063